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QJEEM20Ph3d6b: JEE Main 2020 Physics - Class 11 JEEM20Ph3d6b Solution: Consider two uniform discs of the same thickness and different…

Consider two uniform discs of the same thickness and different radii R1 = R and
R2 = α\alpha R made of the same material. If the ratio of their moments of inertia I1 and I2 , respectively, about their axes is I1 : I2 = 1 : 16 then the value of α\alpha is :

Multiple Choice Options

  • A 2\sqrt 2
  • B 2
  • C 222\sqrt 2
  • D 4

Step-by-Step Verified Solution

Moment of inertia of disc, I=MR22=[p(πR2)t]R22I = {{M{R^2}} \over 2} = {{\left[ {p\left( {\pi {R^2}} \right)t} \right]{R^2}} \over 2}

I=KR4I = K{R^4}

I1I2=(R1R2)4{{{I_1}} \over {{I_2}}} = {\left( {{{{R_1}} \over {{R_2}}}} \right)^4}

116=(RαR)4⇒α=(16)14=2{1 \over {16}} = {\left( {{R \over {\alpha R}}} \right)^4} \Rightarrow \alpha = {\left( {16} \right)^{{1 \over 4}}} = 2

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